Question

In a triangle abc, if (tan a/2)(tan c/2)=1/3 then least value of b is

Answer 1

Step-by-step explanation:

In a triangle

. If

then the least value of

(b) is b9/7

Answer 2

The question is,

If ABC is a triangle such that

$\tan( \frac{a}{2} ) . \tan( \frac{c}{2} ) = \frac{1}{3}$

and a.c=4,Find the least value that b can take

The least value that b can take is 2

Step-by-step explanation:

$\tan( \frac{a}{2} ) = \sqrt{ \frac{(s - b)(s - c)}{s( s - a)} } \\ \tan( \frac{c}{2} ) = \sqrt{ \frac{(s - a)(s - b)}{s( s - c)} }$

such that

$s = \frac{a + b + c}{2}$

$therefore \\ \tan( \frac{a}{2} ) . \tan( \frac{c}{2} ) = \\ \sqrt{ \frac{(s - b)(s - c)}{s( s - a)} } . \sqrt{ \frac{(s - a)(s - b)}{s( s - c)} }$

$= \sqrt{ \frac{(s - b)(s - c)(s - a)(s - b)}{s( s - a)s( s - c)} } \\ = \sqrt{ \frac{ {(s - b)}^{2} }{ {s}^{2} } }$

$=\frac{(s - b)}{s}$

$3(s - b) = s \: implies \: 3s - 3b = s \\ rearranging \: we \: have \: \\ 2s = 3b \\ </p><p>as \: s = \frac{a + b +c }{2}$

$a + b + c = 3b \\ a + c = 2b$

-----equation (1)

$we \: know \: that \: \\ arithemetic \: mean (am)\geqslant geomteric \: mean(gm)</p><p> \\ then \: am = \frac{a + c}{2} \\ and \: gm = \sqrt{ac}$

$\frac{a + c}{2} \geqslant \sqrt{ac \\ }$

using equation 1 in equation above ,we have

$2b \geqslant \sqrt{4} \\ b \geqslant 2$

then the least value that b can have is 2.

DEFINITIONS

$\tan( \frac{a}{2} ) = \\ \sqrt{ \frac{(s - b)(s - c)}{s( s - a)} } .\\ \tan( \frac{b}{2} ) = \\ \sqrt{ \frac{(s - a)(s - c)}{s( s - b)} } . \\ \tan( \frac{c}{2} ) = \\ \sqrt{ \frac{(s - a)(s - b)}{s( s - c)} }$

$s = \frac{a + b + c}{2}$

$\\ arithemetic \: mean (am)\geqslant geomteric \: mean(gm)$

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